I would like to find out the three closest numbers in a vector. Something like
v = c(10,23,25,26,38,50)
c = findClosest(v,3)
c
23 25 26
I tried with sort(colSums(as.matrix(dist(x))))[1:3]
, and it kind of works, but it selects the three numbers with minimum overall distance not the three closest numbers.
There is already an answer for matlab, but I do not know how to translate it to R:
%finds the index with the minimal difference in A
minDiffInd = find(abs(diff(A))==min(abs(diff(A))));
%extract this index, and it's neighbor index from A
val1 = A(minDiffInd);
val2 = A(minDiffInd+1);
How to find two closest (nearest) values within a vector in MATLAB?
Find the closest or nearest number with array formula. For example, you have a list of numbers in Column A, and now you will find the closest value or the nearest value of 18 from the Column A. You can do it as follows: Select a blank cell, and enter below formula, and press the Ctrl + Shift + Enter keys together.
Finding the value in a vector that is closest to a specified value is straightforward using which. Here, we want to find the value of xv that is closest to 108.0: which(abs(xv-108)==min(abs(xv-108))) [1] 332. The closest value to 108.0 is in location 332.
Excel Find Closest Value 1 Select the range where you will search for closest values to the give value, and then click Kutools > Select > Select... 2 In the opening Select Specific Cells dialog box, See More....
Sometimes, you may want to find out and select all closet values to the given value in a range. Actually, we can define a deviation value, and then apply Kutools for Excel’s Select Special Cells utility to find out and select all closest values within the diviation range of give value easily.
My assumption is that the for the n
nearest values, the only thing that matters is the difference between the v[i] - v[i - (n-1)]
. That is, finding the minimum of diff(x, lag = n - 1L)
.
findClosest <- function(x, n) {
x <- sort(x)
x[seq.int(which.min(diff(x, lag = n - 1L)), length.out = n)]
}
findClosest(v, 3L)
[1] 23 25 26
Let's define "nearest numbers" by "numbers with minimal sum of L1 distances". You can achieve what you want by a combination of diff
and windowed sum.
You could write a much shorter function but I wrote it step by step to make it easier to follow.
v <- c(10,23,25,26,38,50)
#' Find the n nearest numbers in a vector
#'
#' @param v Numeric vector
#' @param n Number of nearest numbers to extract
#'
#' @details "Nearest numbers" defined as the numbers which minimise the
#' within-group sum of L1 distances.
#'
findClosest <- function(v, n) {
# Sort and remove NA
v <- sort(v, na.last = NA)
# Compute L1 distances between closest points. We know each point is next to
# its closest neighbour since we sorted.
delta <- diff(v)
# Compute sum of L1 distances on a rolling window with n - 1 elements
# Why n-1 ? Because we are looking at deltas and 2 deltas ~ 3 elements.
withingroup_distances <- zoo::rollsum(delta, k = n - 1)
# Now it's simply finding the group with minimum within-group sum
# And working out the elements
group_index <- which.min(withingroup_distances)
element_indices <- group_index + 0:(n-1)
v[element_indices]
}
findClosest(v, 2)
# 25 26
findClosest(v, 3)
# 23 25 26
A base R option, idea being we first sort
the vector and subtract every i
th element with i + n - 1
element in the sorted vector and select the group which has minimum difference.
closest_n_vectors <- function(v, n) {
v1 <- sort(v)
inds <- which.min(sapply(head(seq_along(v1), -(n - 1)), function(x)
v1[x + n -1] - v1[x]))
v1[inds: (inds + n - 1)]
}
closest_n_vectors(v, 3)
#[1] 23 25 26
closest_n_vectors(c(2, 10, 1, 20, 4, 5, 23), 2)
#[1] 1 2
closest_n_vectors(c(19, 23, 45, 67, 89, 65, 1), 2)
#[1] 65 67
closest_n_vectors(c(19, 23, 45, 67, 89, 65, 1), 3)
#[1] 1 19 23
In case of tie this will return the numbers with smallest value since we are using which.min
.
BENCHMARKS
Since we have got quite a few answers, it is worth doing a benchmark of all the solutions till now
set.seed(1234)
x <- sample(100000000, 100000)
identical(findClosest_antoine(x, 3), findClosest_Sotos(x, 3),
closest_n_vectors_Ronak(x, 3), findClosest_Cole(x, 3))
#[1] TRUE
microbenchmark::microbenchmark(
antoine = findClosest_antoine(x, 3),
Sotos = findClosest_Sotos(x, 3),
Ronak = closest_n_vectors_Ronak(x, 3),
Cole = findClosest_Cole(x, 3),
times = 10
)
#Unit: milliseconds
# expr min lq mean median uq max neval cld
#antoine 148.751 159.071 163.298 162.581 167.365 181.314 10 b
# Sotos 1086.098 1349.762 1372.232 1398.211 1453.217 1553.945 10 c
# Ronak 54.248 56.870 78.886 83.129 94.748 100.299 10 a
# Cole 4.958 5.042 6.202 6.047 7.386 7.915 10 a
An idea is to use zoo
library to do a rolling operation, i.e.
library(zoo)
m1 <- rollapply(v, 3, by = 1, function(i)c(sum(diff(i)), c(i)))
m1[which.min(m1[, 1]),][-1]
#[1] 23 25 26
Or make it into a function,
findClosest <- function(vec, n) {
require(zoo)
vec1 <- sort(vec)
m1 <- rollapply(vec1, n, by = 1, function(i) c(sum(diff(i)), c(i)))
return(m1[which.min(m1[, 1]),][-1])
}
findClosest(v, 3)
#[1] 23 25 26
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